A number when divided successively by 4 and 5 leaves

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Source: Veritas Prep

A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

A. 0
B. 3
C. 4
D. 9
E. 17

The OA is E.

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When we divide a positive integer by another positive integer D (divider), we obtain a quotient Q (which is a non-negative integer) and a remainder R(which is an integer such that 0≤R<D).
Thus we can write n as
n=DQ+r

Now according to our problem ,
x=4Q+1
Now the quotient obtained is divided by 5 yielding a different quotient
Q=5q+4

Substituting the second equation in the first
x=20q +16 +1
=20q +17

Hence, upon dividing the number x by 20 we would get a remainder of 17.

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by fskilnik@GMATH » Sat Oct 13, 2018 2:02 pm
BTGmoderatorLU wrote:Source: Veritas Prep

A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

A. 0
B. 3
C. 4
D. 9
E. 17
$$N = {\rm{original}}\,\,{\rm{number}}$$
$$?\,\,\,:\,\,{\rm{remainder}}\,\,{\rm{of}}\,\,N\,\,{\rm{divided}}\,\,{\rm{by}}\,\,20$$
Explaining the question stem: when the "original number" (N) is divided by 4, the remainder is 1. When the quotient of this division (say Q) is divided by 5, the remainder is 4.

Hence:
$$\left. \matrix{
N = 4Q + 1\,\,\,\left( {Q\,\,{\mathop{\rm int}} } \right) \hfill \cr
Q = 5J + 4\,\,\,\left( {J\,\,{\mathop{\rm int}} } \right) \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\,\,:\,\,N = 4\left( {5J + 4} \right) + 1 = 20J + 17$$
$$? = 17$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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by Scott@TargetTestPrep » Sun Apr 07, 2019 5:12 pm
BTGmoderatorLU wrote:Source: Veritas Prep

A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?

A. 0
B. 3
C. 4
D. 9
E. 17

The OA is E.
We need to find a number that, when divided by 4, leaves a remainder of 1, and when the quotient from this division is divided by 5, a remainder of 4 remains. Let's represent this number by n.

Since our number produces a remainder of 1 when divided by 4, it must be true that n = 4p + 1 for some integer p.

Since the quotient from the previous division, which is p, produces a remainder of 4 when divided by 5, we have p = 5q + 4. Let's substitute this expression of p into the previous equation:

n = 4p + 1

n = 4(5q + 4) + 1

n = 20q + 16 + 1

n = 20q + 17

Finally, since 20q is divisible by 20, the remainder from the division of n by 20 is 17.

Answer: E

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